Q21: Why the current 4G is much faster than the previous generations?
Today, everyone need a connection to the internet. People both need to learn news from the internet and to watch video for entertainment. Considering none wants to wait for minutes to read one slide of news and for hours to buffer a movie. A technology which can provide a high transmission rate is needed on edge. 4G connection is such a technology which can provide a downlink speed up to 1 Gbits/s. Such a high rate can be achieved by implementing MIMO technology. MIMO system is the used widely today to provide a high speed data transmissionTorlak, M.; Duman, T.M., "MIMO communication theory, algorithms, and prototyping," Signal Processing and Communications Applications Conference (SIU), 2012 20th , vol., no., pp.1,2, 18-20 April 2012. The use of multiple transmitting and receiving antennas can provide high spectral efficiency and link reliability for point-to-point communication in fading environments. The difference between SISO and MIMO system compared here is the BER (bit error rate). The lower BER, the better transmission rate. With the same energy constrain, the system with lower BER can achieve a higher transmission rate which we can say the data transmission is faster. =A Short Answer= History of Wireless Communication 1G Cellular system (early 1980s) * 1981 Ericson established NMT (Nordic Mobil Telephone) * 1982 Europe launched GSM (Group Special Mobile) * 1983 AT&T developed the AMPS system (Advanced Mobile Phone Service) Feature * Provide narrow-band voice service. 2.4 kb/s * Based on frequency division multiple access (FDMA). 2G Cellular system (early 1990s) * 1990 In US, the AMPS updated to IS-54/IS-136 based on TDMA * 1992 Qualcomm developed CDMA technology standard IS-95 * 1992 In Europe, TDMA-based GSM spread to world Feature * 2G still targeted narrow-band voice service, short mesage. 9.6 - 14.4 kb/s * Using digital modulation 2.5G Cellular system (around 2000) * GSM had two enhancement ** GPRS (General Packet Radio Service) ** EDGE (Enhanced Data Rates for GSM Evaluation) * US Had CDMA 2000 Feature * Data rate of 164 - 384 kb/s * Voice service and data service 3G Cellular system (early 2000s) * 3G requirements was proposed in standard IMT-2000 (International Mobile Telephone) by IUT (International Telephone Union) * Two major technology ** From Europe, GSM was upgraded to UMTS (Universal Mobile Telephone System) ** From US, CDMA 2000 family 4G Cellular system (around 2010) * Around 2009, 4G system request special in stand IMT-Advanced to support 100Mb/s - 1Gb/s * Two major system ** LTE (Long-Tern Evaluation) 300Mb/s ** WiMax (Worldwide Interoperablily for Microware Access) * MIMO We can see a huge improvement of the data rate in 4G compared with other generations. The reason for this improvement is the 4G wireless communication system takes the advantage of MIMO system. MIMO system model and capacity analyses MIMO system model and ST modulation The MIMO system can be described as the figure. First, the data (i.e. the video or image data) should be first compressed in some special method. Then the compressed data should be converted into bitstream which used in space-time (ST) modulation. After proper modulation scheme, the signal is ready to be transmitted. Before the transmission, the ST signal should be timed a constant \sqrt{\frac{\rho}{M_t}} which have some thing to do with the energy normalization. In ST modulation, we select one matrix from a set of matrix (constellation). Each ST signal is a T by M_t matrix. The form of the signal is shown as following, \mathbf{C} = \left( \begin{array}{cccc} C_1^1 & C_1^2 & \ldots & C_1^{M_t} \\ C_2^2 & C_2^2 & \ldots & C_2^{M_t} \\ \vdots & \vdots & \ddots & \vdots \\ C_T^1 & C_T^2 & \ldots & C_T^{M_t} \\ \end{array} \right) Where, each row indicates the symbols sent at the same time slot by all the antennas and each column indicates the symbols sent by the same antenna at all the time slots. The entries of \mathbf{C} can be arbitrary complex number with the power normalization is ||\mathbf{C}||_F^2 = T \times M_t . MIMO capacity Based on the system shown in the figure. The MIMO system capacity can be obtained as C = \log \det (\mathbf{I}_{M_t} + \frac{\rho}{M_t}\mathbf{H}_{H} \mathbf{H}) . Where \rho is the average SNR. \mathbf{I}_{M_t} is the identity matrix of size M_r by M_t . Some research deals with the MIMO capacity has been postedGoldsmith, A.; Jafar, S.A.; Jindal, N.; Vishwanath, S., "Capacity limits of MIMO channels," Selected Areas in Communications, IEEE Journal on , vol.21, no.5, pp.684,702, June 2003Uthansakul, P.; Bialkowski, M.E., "Multipath signal effect on the capacity of MIMO, MIMO-OFDM and spread MIMO-OFDM," Microwaves, Radar and Wireless Communications, 2004. MIKON-2004. 15th International Conference on , vol.3, no., pp.989,992 Vol.3, 17-19 May 2004. We can see that MIMO system has a much smaller error probability than the SISO system. So, MIMO system has a large capacity than SISO system. =A Long Answer= MIMO Transceiver Signal Model See the figure which shows the system model, where * C^i_t : The signal transmitted by Tx (transmitter) i at time slot t . * y^i_t : Received signal at Rx (receiver) i at time slot t . * h_{i,j} : Channel coefficient from Tx i to Rx j . ** Assume that the channel is flat Rayleigh fading. ** Assume h_{i,j} \in \mathcal{CN}(0,1) and they are independent to each other. ** Assume that the channel is quasi-static i.e. the channel coefficients do not change during one block transmission, and they may change independently from one block to another. At time slot t , the signal received signal at Rx j is given by: y_t^j = \sqrt{\frac{\rho}{M_t}} \sum_{i=1}^{M_t}h_{i,j}C_t^i+\eta_t^j where j \in (1,2, \dots, M_r) , \eta_t^j is a zero-mean complex Gaussian variable with variance 1 (i.e. \eta_t^j \in \mathcal{CN}(0,1) ). Then, we can rewrite the transceiver signal in a compact matrix form: \mathbf{Y} = \sqrt{\frac{\rho}{M_t}} \mathbf{C H} + \mathbf{N} where, * \mathbf{Y} : T \times M_r matrix with the form \mathbf{Y} = \left( \begin{array}{cccc} y_1^1 & y_1^2 & \ldots & y_1^{M_r} \\ y_2^2 & y_2^2 & \ldots & y_2^{M_r} \\ \vdots & \vdots & \ddots & \vdots \\ y_T^1 & y_T^2 & \ldots & y_T^{M_r} \\ \end{array} \right) to represent the received signal matrix. * \mathbf{H} : M_t \times M_r matrix with the form \mathbf{H} = \left( \begin{array}{cccc} h_1^1 & h_1^2 & \ldots & h_1^{M_r} \\ h_2^2 & h_2^2 & \ldots & h_2^{M_r} \\ \vdots & \vdots & \ddots & \vdots \\ h_{M_t}^1 & h_{M_t}^2 & \ldots & h_{M_t}^{M_r} \\ \end{array} \right) to represent the channel coefficients. * \mathbf{N} : T \times M_r matrix with the form \mathbf{N} = \left( \begin{array}{cccc} \eta_1^1 & \eta_1^2 & \ldots & \eta_1^{M_r} \\ \eta_2^2 & \eta_2^2 & \ldots & \eta_2^{M_r} \\ \vdots & \vdots & \ddots & \vdots \\ \eta_T^1 & \eta_T^2 & \ldots & \eta_T^{M_r} \\ \end{array} \right) to represent the noise. * \mathbf{C} : T \times M_r matrix with the form shown above. ST signal Decoding and Demodulation Suppose that MIMO system uses a set of ST signal {C_0, C_1, \dots, C_{L-1}} . If a ST signal C_{L0} is sent out by assuming the channel information \mathbf{H} is known at Rx, then the Maximum Likelihood (ML) demodulation is \hat{C} = \arg \min_{C_l \in {C_0, \dots, C_{L-1}}} || \mathbf{Y} - \sqrt{\frac{\rho}{M_t}} C_l \mathbf{H} + \mathbf{N}||_F^2 Preference Analysis Suppose that we have a set of ST signal {C_0, C_1, \dots, C_{L-1}} and each of them has equal probability to be sent out. Then the probability of incorrect decoding at Rx is Q(\sqrt{\frac{\rho}{M_t}}||(C - \tilde{C})\mathbf{H}||_F) , where Q(.) is the Q function. The figure shows the result of the comparison between simulation PEP (Pairwise Error Probability) and theoretical PEP. ST signal Design Criteria Based on the PEP analysis, two design criteria were proposed * Rank Criterion or Diversity Criterion: The minimum rank of the difference matrix C-\tilde{C} over all pairs of the different codeword \tilde{C} should be as large as possible. * Produce Criterion: The minimum value of the product \prod_{i=1}^r{\lambda_i} ( \lambda_i is the eigenvalue) over all distinct signals C and \tilde{C} should be as large as possible. Some designs are based on the channel conditionsNabar, R.U.; Bolcskei, H., "Space-time signal design for fading relay channels," Global Telecommunications Conference, 2003. GLOBECOM '03. IEEE , vol.4, no., pp.1952,1956 vol.4, 1-5 Dec. 2003Su, W.; Safar, Z.; Liu, K. J R, "Space-time signal design for time-correlated Rayleigh fading channels," Communications, 2003. ICC '03. IEEE International Conference on , vol.5, no., pp.3175,3179 vol.5, 11-15 May 2003. Some well-known ST signals Goal: * ||C_l||_F^2 = M^2_t * The difference matrix between any two distinct codewords C_l and C_L^' should be full rank (diversity). * The coding gain \xi should be as large as possible. \xi = \frac{1}{2 \sqrt{M_t}} \min_{l \ne l^'} |\det (C_l - C_l^')|^{\frac{1}{M_t}} Cyclic Space-Time Block Codes The signals have the form as C_l = \left( \begin{array}{cccc} e^{ju_1 \theta_l} & 0 & \ldots & 0 \\ 0 & e^{ju_2 \theta_l} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & e^{ju_{M_t} \theta_l} \\ \end{array} \right) where \theta=\frac{l}{L}2\pi , l = 0, 1, \dots, L-1 and u_1, u_2, \dots, u_{M_t} \in {0, 1, \dots, L-1} are some integers to be optimized. The cyclic STBC can achieve full diversityHong-Liang, Fu; Feng Guang-zeng, "Cyclic Space-Time Block Codes and Decoding Algorithm in MIMO CDMA System," Wireless Communications, Networking and Mobile Computing, 2006. WiCOM 2006.International Conference on , vol., no., pp.1,4, 22-24 Sept. 2006Liang Xian; Huaping Liu, "Space-time block codes from cyclic design," Communications Letters, IEEE , vol.9, no.3, pp.231,233, March 2005. Calculate the coding gain of Cyclic STBC \xi = \frac{1}{2 \sqrt{M_t}} \min_{l \ne l^'} |\det (C_l - C_l^')|^{\frac{1}{M_t}} \quad = \frac{1}{2 \sqrt{M_t}} \min_{l \ne l^'} |(\sqrt{M_t}^{M_t} \prod_{i=1}^{M_t} |2 \sin \frac{u_i \theta_{l = l^'}}{2}|)|^{\frac{1}{M_t}} \quad = \min_{l \ne l^'} \prod_{i=1}^{M_t} |\sin \frac{u_i \theta_{l = l^'}}{2}| ^{\frac{1}{M_t}} = \min_{1 \leqslant m \leqslant L-1} \prod_{i=1}^{M_t} |\sin \frac{u_i m \pi}{L}|^{\frac{1}{M_t}} Unitary Space-Time Block Codes Each M_r \times M_t ST signal comes from unitary matrices such that \left( \frac{1}{\sqrt{M_t}} \right)^H \left(\frac{1}{\sqrt{M_t}}\right) = I_{M_r \times M_t} The cyclic STBC is one of unitary STBC is one of unitary STBC. There are other uniary STBC that are better than the cyclic code. This kind of codes design can be combined with the differential design which will be discussed in following chaptersHimsoon, T.; Su, W.; Liu, K. J R, "Differential unitary space-time signal design using matrix rotation structure," Signal Processing Letters, IEEE , vol.12, no.1, pp.45,48, Jan. 2005. Some other novel designs are also postedSmith, D.B.; Hanlen, L.W., "Novel Unitary Space-Time Signal Design," Information Theory Workshop, 2006. ITW '06 Punta del Este. IEEE , vol., no., pp.327,331, 13-17 March 2006Allen, B.; Kuroda, Y.; Said, F.; Aghvami, A.H., "Performance of differential and differential unitary space-time block codes over spatially correlated Rayleigh fading channels," 3G Mobile Communication Technologies, 2004. 3G 2004. Fifth IEE International Conference on , vol., no., pp.378,382, 2004. Orthogonal Space-Time Block Codes Orthogonal STBC has to advantages, * Achieve full diversity * They have simple fast ML decoding (symbols can be decoded individually) Some kinds of orthogonal STBC has been posted such as Quasi-Orthogonal Space-Time Block Codes with full diversity Su, W. and Xia, Xiang-Gen, "A design of quasi-orthogonal space-time block codes with full diversity," Signals, Systems and Computers, 2002. Conference Record of the Thirty-Sixth Asilomar Conference, vol. 2, pp. 1112-1116, 3-6 Nov. 2002 Some works were done based on differential transceiver design for 1uasi-Orthogonal Space-Time Block CodesLingyang Song; Hjorungnes, A.; Burr, A.G.; Bhatnagar, M., "Differential Transceiver Design for Quasi-Orthogonal Space-Time Block Codes," Wireless Communications, Networking and Mobile Computing, 2007. WiCom 2007. International Conference on , vol., no., pp.283,286, 21-25 Sept. 2007. A motivated example -- Alamonti scheme Alamonti (1998) proposed a simple scheme for systems with two Tx as follows G_2(x_1, x_2) = \left( \begin{array}{cc} x_1 & x_2\\ -x_2^* & x_1^*\\ \end{array} \right) where x_1 and x_2 are arbitrary complex symbols. This kind of STBC has very simple fast ML decoding. The information symbols can be decoded separately not jointly. This kind of separation can largely decreasing the decoding complexity. Diagonal Algebraic ST codes For k information symbols: S_1, S_2, \dots, S_k , we define x_2, \dots, x_k \triangleq S_2, \dots, S_k V(\theta_1, \theta_2, \dots, \theta_k) . Where V(\dot) is a Vandermonde matrix with varibles \theta_1, \theta_2, \dots, \theta_k with the following form, V(\theta_1, \theta_2, \dots, \theta_k) = \left( \begin{array}{cccc} 1 & 1 & \cdots & 1\\ \theta_1^1 & \theta_2^1 & \cdots & \theta_k^1\\ \vdots & \vdots & \ddots & \vdots \\ \theta_1^{k-1} & \theta_2^{k-1} & \cdots & \theta_k^{k-1}\\ \end{array} \right)_{k \times k} Then, from a diagonal ST codewords as following, C = \left( \begin{array}{cccc} x_1 & 0 & \cdots & 0\\ 0 & x_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & x_k\\ \end{array} \right) . This kind of signal can also achieve full diversityDamen, M.-O.; Abed-Meraim, K.; Belfiore, J. -C, "Diagonal algebraic space-time block codes," Information Theory, IEEE Transactions on , vol.48, no.3, pp.628,636, Mar 2002. =Examples= Optimal Cyclic STBC Coding Gain For example, for M_t = 2 and L = 4 , we can find the optimal u to obtain the largest coding gain. The result is \quad u_2 = \quad 1 . Thus, the resulting cyclic ST signals are given by C_j = \sqrt{2} \left( \begin{array}{cc} e^{j\theta_l} & 0 \\ 0 & e^{j\theta_l} \end{array} \right) \qquad \theta_l = \frac{l \pi}{2} \qquad l = 0,1,2,3 The normalize coding gian \xi = \min_{1 \leqslant m \leqslant 3} \prod_{i=1}^{2} |\sin \frac{u_i m \pi}{4}|^{\frac{1}{2}} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} Problem: For example, for M_t = 2 and L = 16 , we can find the optimal u to obtain the largest coding gain. The result is \quad u_2 = \quad 7 . Thus, the resulting cyclic ST signals are given by C_j = \sqrt{2} \left( \begin{array}{cc} e^{j\theta_l} & 0 \\ 0 & e^{j7\theta_l} \end{array} \right) \qquad \theta_l = \frac{l \pi}{8} \qquad l = 0,1,\dots,15 Fast Decoding Scheme Please prove this kind of scheme can achieve fast Maximum Likelihood decoding (information symbols can be decoded separately not jointly). Solution: Assume that the channel H is know at the Rx then the ML decoding is (\hat{x_1},\hat{x_2}) = \arg \min_{x_1,x_2} || Y - \sqrt{\frac{\rho}{2}} G_2(x_1,x_2) H ||_F^2 We donate G_2(x_1,x_2) as G_2 and use ||A+B||^2_F = tr\{(A+B)^H(A+B)\} = tr\{A^H A\}+2Re\{tr(A^HB) \}+tr\{B^HB\} to separate the equation, =\arg \min_{x_1,x_2} (tr\{Y^H Y\} - 2Re\{tr(Y^H \sqrt{\frac{\rho}{2}} G_2 H)\} + \frac{\rho}{2} tr \{(G_2 H)^H G_2 H\}) Then, we plug in the code word matrix of G_2 , = \arg \min_{x_1,x_2} (||Y||^2_F - \sqrt{2 \rho} Re\{tr(Y^H G_2 H)\} + \frac{\rho}{2} tr \{H^* (||x_1||^2 + ||x_2||^2) I_{2 \times 2} H \}) = \arg \min_{x_1,x_2} (||Y||^2_F - \sqrt{2 \rho} Re\{tr(Y^H G_2 H)\} + \frac{\rho}{2} (||x_1||^2 + ||x_2||^2) ||H||_F^2) Because we know channel H and received signal Y , we can donate HY^H as a constant \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] As a result, tr(G_2 HY^H) = tr( \left[ \begin{array}{cc} x_1 & x_2 \\ -x_2^* & x_1^* \end{array} \right] \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]) = ax_1 + cx_2 - bx_2^* + dx_1^* (\hat{x_1},\hat{x_2}) = \arg \min_{x_1,x_2} (-\sqrt{2\rho}Re(ax_1 + dx_1^*) + \frac{\rho}{2} ||x_1||^2 ||H||_F^2 - \sqrt{2\rho}Re(cx_2 - bx_2^*) + \frac{\rho}{2} ||x_2||^2 ||H||_F^2) We can donate f_1(x_1) = -\sqrt{2\rho}Re(ax_1 + dx_1^*) + \frac{\rho}{2} ||x_1||^2 ||H||_F^2 and f_2(x_2) = - \sqrt{2\rho}Re(cx_2 - bx_2^*) + \frac{\rho}{2} ||x_2||^2 ||H||_F^2) (\hat{x_1},\hat{x_2}) = \arg \min_{x_1,x_2} (f_1(x_1) + f_2(x_2)) = (\arg \min_{x_1}f_1(x_1), \arg \min_{x_2}f_2(x_2)) In this way, we can decode these two symbols separately by searching only 2L different combinations instead of searching L \times L combinations. For example, if we use 16-QAM, the Alamouti scheme only tests 32 different combinations instead of 256 combinations. Maximum Likelihood Detection If we want to know what symbol is transmitted in current time slot, the decode should use the Maximum Likelihood Detection to estimate the transmitted symbol. We use the following process to detect the transmitted symbols. \hat{C} = \arg \min_{C_l \in {C_0, \dots, C_{L-1}}} || \mathbf{Y} - \sqrt{\frac{\rho}{M_t}} C_l \mathbf{H}||_F^2 Where Y is the received signal given by, \mathbf{Y} = \sqrt{\frac{\rho}{M_t}} \mathbf{C H} + \mathbf{N} First, we assume the channel is known at the receivers and is modeled as complex Gaussian Random Variables with mean zero and variance one. The noise is also modeled as complex Gaussian Random Variables with mean zero and variance one. The value of channel and noise is shown as following, h_1 = \left[ \begin{array}{c} 0.5377 - 2.2588i \\ 1.8339 + 0.8622i \end{array} \right] h_2 = \left[ \begin{array}{c} 0.3188 - 0.4336i \\ -1.3077 + 0.3426i \end{array} \right] Noise, n_1 = \left[ \begin{array}{c} 3.5784 - 1.3499i \\ 2.7694 + 3.0349i \end{array} \right] n_2 = \left[ \begin{array}{c} 0.7254 + 0.7147i \\ -0.0631 - 0.2050i \end{array} \right] The information symbols we use is, C_0 = \left[ \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right] C_1 = \left[ \begin{array}{cc} -1 & -1 \\ 1 & -1 \end{array} \right] Other parameters M_t = 2 and \rho = 10 . We assume at first time slot we transmitted symbol C_1 and in time second time slot we transmitted symbol C_0 . In the first time slot, we first simulate the received signal though the channel, \mathbf{Y_1} = \sqrt{\frac{10}{2}} \left[ \begin{array}{cc} -1 & -1 \\ 1 & -1 \end{array} \right] \left[ \begin{array}{c} 3.5784 - 1.3499i \\ 2.7694 + 3.0349i \end{array} \right] + \left[ \begin{array}{cc} 0.7254 + 0.7147i \\ -0.0631 - 0.2050i \end{array} \right] = \left[ \begin{array}{cc} -1.7247 + 1.7730i\\ -0.1290 - 3.9439i \end{array} \right] Calculate || \mathbf{Y_1} - \sqrt{\frac{\rho}{M_t}} C_l \mathbf{H}||_F^2 for both two possible symbols C_0 and C_1 . The result is 14.2066 and 5.6131 for symbols C_0 and C_1 accordingly. The smallest value is the value according to C_1 , so, the result of detection is C_1 . The detection get the right result. We can do the same in the second time slot, the result is still a correct detection. But, please notice, the detection result is not always right. What's the reason and at what kind of situation will dead to a fault detection. Answer: The reason is the noise will do harm to the process of detection. Usually at the time slot which channel condition is bad (i.e. the channel coefficient is weak in energy) and the noise is strong. The detection will fail. This kind of process can be repeated many times to get a curve like shown in the plot. By doing this, we can learn that MIMO system can achieve a better detection than SISO system. =Advanced Material= MIMO Transmission and Detection There are three scenarios in MIMO communication depends on the knowledge of Channel State Information (CSI), * CSI is available at Rx, not Tx * CSI is not available at neither Rx nor Tx * CSI is available at both Rx and Tx CSI is available at Rx, not Tx If the CSI is available at Rx, not Tx, we could implement the ML (Maximum Likelihood) detection which has shown in the above introduction. CSI is not available at neither Rx nor Tx In this scenario, we should implement non-coherent detection and differentical ST modulation and demodulation scheme for MIMO system. The main consideration for doing this is to try to cancel the channel information by using the adjacent transmission. We can do this if the channel coefficients are approximately the same over two adjacent transmission blocks (i.e. H_{\tau} \approx H_{\tau-1} ). This kind of scheme can be formed as following Y_{\tau} = \sqrt{\frac{1}{M_t}} \mathcal{S}_{\tau} \mathbf{H}_{\tau} + \mathbf{N}_{\tau} where \mathcal{S}_{\tau} is the transmitted matrix at time \tau . As a result, consider two adjacent transmission sections, Y_{\tau} = \sqrt{\frac{1}{M_t}} \mathcal{S}_{\tau} \mathbf{H}_{\tau} + \mathbf{N}_{\tau} Y_{\tau-1} = \sqrt{\frac{1}{M_t}} \mathcal{S}_{\tau-1} \mathbf{H}_{\tau-1} + \mathbf{N}_{\tau-1} Because we already know the received signal at each time slot (i.e. Y_{\tau} and Y_{\tau-1} ), we can formulate the ML detection as following, \hat{C_{l,\tau}} = \arg \min_{l = 1,2,\dots,L-1} ||Y_{\tau} - \frac{1}{\sqrt{M_t}}C_{l,\tau}Y_{\tau-1}||_F^2 CSI is available at both Rx and Tx If both Rx and Tx know the CSI, we should apply a transmit beamforming to take this advantage. =References=